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How to find the largest and smallest values of a function in a bounded closed area? Investigation of the graph of a function.

In this article I will talk about how to apply the ability to find to the study of a function: to find its largest or the smallest value. And then we will solve some problems from Task B15 from open bank assignments for .

As usual, let's start with the theory first.

At the beginning of any study of a function, we find it

To find the largest or smallest value of the function, you need to investigate on which intervals the function increases and on which it decreases.

To do this, you need to find the derivative of the function and study its intervals of constant sign, that is, the intervals on which the derivative retains its sign.

The intervals on which the derivative of a function is positive are intervals of increasing function.

The intervals on which the derivative of a function is negative are intervals of decreasing function.

1 . Let's solve task B15 (No. 245184)

To solve it, we will follow the following algorithm:

a) Find the domain of the function

b) Find the derivative of the function .

c) Set it equal to zero.

d) Let us find the intervals of constant sign of the function.

e) Find the point at which the function takes the greatest value.

f) Find the value of the function at this point.

I tell the detailed solution of this task in the VIDEO LESSON:

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2. Let's solve task B15 (No. 282862)

Find the largest value of a function on the segment

It is obvious that the function takes the greatest value on the segment at the maximum point, at x=2. Find the value of the function at this point:

Answer: 5

3 . Let's solve task B15 (No. 245180):

Find the largest value of a function

1.title="ln5>0">, , т.к. title="5>1">, поэтому это число не влияет на знак неравенства.!}

2. Since the scope of the original function title="4-2x-x^2>0">, следовательно знаменатель дроби всегда больще нуля и дробь меняет знак только в нуле числителя.!}

3. The numerator is zero at . Let's check if the ODZ belongs to the function. To do this, check if the condition title="4-2x-x^2>0"> при .!}

Title="4-2(-1)-((-1))^2>0">,

so the point belongs to the ODZ of the function

We examine the sign of the derivative to the right and left of the point:

We see that the function takes the greatest value at the point . Now let's find the value of the function at :

Note 1. Note that in this problem we did not find the domain of the function: we only fixed the constraints and checked whether the point at which the derivative is equal to zero belongs to the domain of the function. In this problem, this turned out to be enough. However, this is not always the case. It depends on the task.

Remark 2. When studying the behavior of a complex function, one can use the following rule:

if the outer function of a compound function is increasing, then the function takes on its greatest value at the same point at which the inner function takes on its greatest value. This follows from the definition of an increasing function: a function increases on the interval I if greater value an argument from this interval corresponds to a larger value of the function.
if the outer function of a complex function is decreasing, then the function takes on the largest value at the same point at which the inner function takes on the smallest value . This follows from the definition of a decreasing function: the function decreases on the interval I if the larger value of the argument from this interval corresponds to the smaller value of the function

In our example, the outer function - increases over the entire domain of definition. Under the sign of the logarithm is an expression - a square trinomial, which, with a negative senior coefficient, takes the largest value at the point . Next, we substitute this value of x into the equation of the function and find its largest value.

Let the function $z=f(x,y)$ be defined and continuous in some bounded closed domain $D$. Let in this area for given function has finite partial derivatives of the first order (with the possible exception of a finite number of points). To find the largest and smallest values of a function of two variables in a given closed region, three steps of a simple algorithm are required.

Algorithm for finding the largest and smallest values of the function $z=f(x,y)$ in the closed domain $D$.

Find the critical points of the function $z=f(x,y)$ that belong to the region $D$. Compute function values at critical points.
Investigate the behavior of the function $z=f(x,y)$ on the boundary of the region $D$ by finding the points of possible maximum and minimum values. Calculate the function values at the obtained points.
From the function values obtained in the previous two paragraphs, choose the largest and smallest.

What are critical points? show/hide

Under critical points imply points where both first-order partial derivatives are equal to zero (i.e. $\frac(\partial z)(\partial x)=0$ and $\frac(\partial z)(\partial y)=0 $) or at least one partial derivative does not exist.

Often the points at which the first-order partial derivatives are equal to zero are called stationary points. Thus, stationary points are a subset of critical points.

Example #1

Find the maximum and minimum values of the function $z=x^2+2xy-y^2-4x$ in the closed region bounded by the lines $x=3$, $y=0$ and $y=x+1$.

We will follow the above, but first we will deal with the drawing of a given area, which we will denote by the letter $D$. We are given the equations of three straight lines, which limit this area. The straight line $x=3$ passes through the point $(3;0)$ parallel to the y-axis (axis Oy). The straight line $y=0$ is the equation of the abscissa axis (Ox axis). Well, to construct a straight line $y=x+1$ let's find two points through which we draw this straight line. You can, of course, substitute a couple of arbitrary values instead of $x$. For example, substituting $x=10$, we get: $y=x+1=10+1=11$. We have found the point $(10;11)$ lying on the line $y=x+1$. However, it is better to find those points where the line $y=x+1$ intersects with the lines $x=3$ and $y=0$. Why is it better? Because we will lay down a couple of birds with one stone: we will get two points for constructing the straight line $y=x+1$ and at the same time find out at what points this straight line intersects other lines that bound the given area. The line $y=x+1$ intersects the line $x=3$ at the point $(3;4)$, and the line $y=0$ - at the point $(-1;0)$. In order not to clutter up the course of the solution with auxiliary explanations, I will put the question of obtaining these two points in a note.

How were the points $(3;4)$ and $(-1;0)$ obtained? show/hide

Let's start from the point of intersection of the lines $y=x+1$ and $x=3$. The coordinates of the desired point belong to both the first and second lines, so to find unknown coordinates, you need to solve the system of equations:

$$ \left \( \begin(aligned) & y=x+1;\\ & x=3. \end(aligned) \right. $$

The solution of such a system is trivial: substituting $x=3$ into the first equation we will have: $y=3+1=4$. The point $(3;4)$ is the desired intersection point of the lines $y=x+1$ and $x=3$.

Now let's find the point of intersection of the lines $y=x+1$ and $y=0$. Again, we compose and solve the system of equations:

$$ \left \( \begin(aligned) & y=x+1;\\ & y=0. \end(aligned) \right. $$

Substituting $y=0$ into the first equation, we get: $0=x+1$, $x=-1$. The point $(-1;0)$ is the desired intersection point of the lines $y=x+1$ and $y=0$ (abscissa axis).

Everything is ready to build a drawing that will look like this:

The question of the note seems obvious, because everything can be seen from the figure. However, it is worth remembering that the drawing cannot serve as evidence. The figure is just an illustration for clarity.

Our area was set using the equations of lines that limit it. It's obvious that these lines define a triangle, don't they? Or not quite obvious? Or maybe we are given a different area, bounded by the same lines:

Of course, the condition says that the area is closed, so the picture shown is wrong. But to avoid such ambiguities, it is better to define regions by inequalities. We are interested in the part of the plane located under the line $y=x+1$? Ok, so $y ≤ x+1$. Our area should be located above the line $y=0$? Great, so $y ≥ 0$. By the way, the last two inequalities are easily combined into one: $0 ≤ y ≤ x+1$.

$$ \left \( \begin(aligned) & 0 ≤ y ≤ x+1;\\ & x ≤ 3. \end(aligned) \right. $$

These inequalities define the domain $D$, and define it uniquely, without any ambiguities. But how does this help us in the question at the beginning of the footnote? It will also help :) We need to check if the point $M_1(1;1)$ belongs to the region $D$. Let us substitute $x=1$ and $y=1$ into the system of inequalities that define this region. If both inequalities are satisfied, then the point lies inside the region. If at least one of the inequalities is not satisfied, then the point does not belong to the region. So:

$$ \left \( \begin(aligned) & 0 ≤ 1 ≤ 1+1;\\ & 1 ≤ 3. \end(aligned) \right. \;\; \left \( \begin(aligned) & 0 ≤ 1 ≤ 2;\\ & 1 ≤ 3. \end(aligned) \right.$$

Both inequalities are true. The point $M_1(1;1)$ belongs to the region $D$.

Now it is the turn to investigate the behavior of the function on the boundary of the domain, i.e. go to. Let's start with the straight line $y=0$.

The straight line $y=0$ (abscissa axis) limits the region $D$ under the condition $-1 ≤ x ≤ 3$. Substitute $y=0$ into the given function $z(x,y)=x^2+2xy-y^2-4x$. The resulting substitution function of one variable $x$ will be denoted as $f_1(x)$:

$$ f_1(x)=z(x,0)=x^2+2x\cdot 0-0^2-4x=x^2-4x. $$

Now for the function $f_1(x)$ we need to find the largest and smallest values on the interval $-1 ≤ x ≤ 3$. Find the derivative of this function and equate it to zero:

$$ f_(1)^(")(x)=2x-4;\\ 2x-4=0; \; x=2. $$

The value $x=2$ belongs to the segment $-1 ≤ x ≤ 3$, so we also add $M_2(2;0)$ to the list of points. In addition, we calculate the values of the function $z$ at the ends of the segment $-1 ≤ x ≤ 3$, i.e. at the points $M_3(-1;0)$ and $M_4(3;0)$. By the way, if the point $M_2$ did not belong to the segment under consideration, then, of course, there would be no need to calculate the value of the function $z$ in it.

So, let's calculate the values of the function $z$ at the points $M_2$, $M_3$, $M_4$. You can, of course, substitute the coordinates of these points in the original expression $z=x^2+2xy-y^2-4x$. For example, for the point $M_2$ we get:

$$z_2=z(M_2)=2^2+2\cdot 2\cdot 0-0^2-4\cdot 2=-4.$$

However, the calculations can be simplified a bit. To do this, it is worth remembering that on the segment $M_3M_4$ we have $z(x,y)=f_1(x)$. I'll spell it out in detail:

\begin(aligned) & z_2=z(M_2)=z(2,0)=f_1(2)=2^2-4\cdot 2=-4;\\ & z_3=z(M_3)=z(- 1,0)=f_1(-1)=(-1)^2-4\cdot (-1)=5;\\ & z_4=z(M_4)=z(3,0)=f_1(3)= 3^2-4\cdot 3=-3. \end(aligned)

Of course, there is usually no need for such detailed entries, and in the future we will begin to write down all calculations in a shorter way:

$$z_2=f_1(2)=2^2-4\cdot 2=-4;\; z_3=f_1(-1)=(-1)^2-4\cdot (-1)=5;\; z_4=f_1(3)=3^2-4\cdot 3=-3.$$

Now let's turn to the straight line $x=3$. This line bounds the domain $D$ under the condition $0 ≤ y ≤ 4$. Substitute $x=3$ into the given function $z$. As a result of such a substitution, we get the function $f_2(y)$:

$$ f_2(y)=z(3,y)=3^2+2\cdot 3\cdot y-y^2-4\cdot 3=-y^2+6y-3. $$

For the function $f_2(y)$, you need to find the largest and smallest values on the interval $0 ≤ y ≤ 4$. Find the derivative of this function and equate it to zero:

$$ f_(2)^(")(y)=-2y+6;\\ -2y+6=0; \; y=3. $$

The value $y=3$ belongs to the segment $0 ≤ y ≤ 4$, so we add $M_5(3;3)$ to the points found earlier. In addition, it is necessary to calculate the value of the function $z$ at the points at the ends of the segment $0 ≤ y ≤ 4$, i.e. at the points $M_4(3;0)$ and $M_6(3;4)$. At the point $M_4(3;0)$ we have already calculated the value of $z$. Let us calculate the value of the function $z$ at the points $M_5$ and $M_6$. Let me remind you that on the segment $M_4M_6$ we have $z(x,y)=f_2(y)$, therefore:

\begin(aligned) & z_5=f_2(3)=-3^2+6\cdot 3-3=6; &z_6=f_2(4)=-4^2+6\cdot 4-3=5. \end(aligned)

And, finally, consider the last boundary of $D$, i.e. line $y=x+1$. This line bounds the region $D$ under the condition $-1 ≤ x ≤ 3$. Substituting $y=x+1$ into the function $z$, we will have:

$$ f_3(x)=z(x,x+1)=x^2+2x\cdot (x+1)-(x+1)^2-4x=2x^2-4x-1. $$

Once again we have a function of one variable $x$. And again, you need to find the largest and smallest values of this function on the segment $-1 ≤ x ≤ 3$. Find the derivative of the function $f_(3)(x)$ and equate it to zero:

$$ f_(3)^(")(x)=4x-4;\\ 4x-4=0; \; x=1. $$

The value $x=1$ belongs to the interval $-1 ≤ x ≤ 3$. If $x=1$, then $y=x+1=2$. Let's add $M_7(1;2)$ to the list of points and find out what the value of the function $z$ is at this point. The points at the ends of the segment $-1 ≤ x ≤ 3$, i.e. points $M_3(-1;0)$ and $M_6(3;4)$ were considered earlier, we have already found the value of the function in them.

$$z_7=f_3(1)=2\cdot 1^2-4\cdot 1-1=-3.$$

The second step of the solution is completed. We got seven values:

$$z_1=-2;\;z_2=-4;\;z_3=5;\;z_4=-3;\;z_5=6;\;z_6=5;\;z_7=-3.$$

Let's turn to. Choosing the largest and smallest values from those numbers that were obtained in the third paragraph, we will have:

$$z_(min)=-4; \; z_(max)=6.$$

The problem is solved, it remains only to write down the answer.

Answer: $z_(min)=-4; \; z_(max)=6$.

Example #2

Find the largest and smallest values of the function $z=x^2+y^2-12x+16y$ in the region $x^2+y^2 ≤ 25$.

Let's build a drawing first. The equation $x^2+y^2=25$ (this is the boundary line of the given area) defines a circle with a center at the origin (i.e. at the point $(0;0)$) and a radius of 5. The inequality $x^2 +y^2 ≤ 25$ satisfy all points inside and on the mentioned circle.

We will act on. Let's find partial derivatives and find out the critical points.

$$ \frac(\partial z)(\partial x)=2x-12; \frac(\partial z)(\partial y)=2y+16. $$

There are no points at which the found partial derivatives do not exist. Let us find out at what points both partial derivatives are simultaneously equal to zero, i.e. find stationary points.

$$ \left \( \begin(aligned) & 2x-12=0;\\ & 2y+16=0. \end(aligned) \right. \;\; \left \( \begin(aligned) & x =6;\\ & y=-8.\end(aligned) \right.$$

We got a stationary point $(6;-8)$. However, the found point does not belong to the region $D$. This is easy to show without even resorting to drawing. Let's check if the inequality $x^2+y^2 ≤ 25$, which defines our domain $D$, holds. If $x=6$, $y=-8$, then $x^2+y^2=36+64=100$, i.e. the inequality $x^2+y^2 ≤ 25$ is not satisfied. Conclusion: the point $(6;-8)$ does not belong to the region $D$.

Thus, there are no critical points inside $D$. Let's move on, to. We need to investigate the behavior of the function on the boundary of the given area, i.e. on the circle $x^2+y^2=25$. You can, of course, express $y$ in terms of $x$, and then substitute the resulting expression into our function $z$. From the circle equation we get: $y=\sqrt(25-x^2)$ or $y=-\sqrt(25-x^2)$. Substituting, for example, $y=\sqrt(25-x^2)$ into the given function, we will have:

$$ z=x^2+y^2-12x+16y=x^2+25-x^2-12x+16\sqrt(25-x^2)=25-12x+16\sqrt(25-x ^2); \;\; -5≤ x ≤ 5. $$

The further solution will be completely identical to the study of the behavior of the function on the boundary of the region in the previous example No. 1. However, it seems to me more reasonable in this situation to apply the Lagrange method. We are only interested in the first part of this method. After applying the first part of the Lagrange method, we will get points at which and examine the function $z$ for the minimum and maximum values.

We compose the Lagrange function:

$$ F=z(x,y)+\lambda\cdot(x^2+y^2-25)=x^2+y^2-12x+16y+\lambda\cdot (x^2+y^2 -25). $$

We find the partial derivatives of the Lagrange function and compose the corresponding system of equations:

$$ F_(x)^(")=2x-12+2\lambda x; \;\; F_(y)^(")=2y+16+2\lambda y.\\ \left \( \begin (aligned) & 2x-12+2\lambda x=0;\\ & 2y+16+2\lambda y=0;\\ & x^2+y^2-25=0.\end(aligned) \ right. \;\; \left \( \begin(aligned) & x+\lambda x=6;\\ & y+\lambda y=-8;\\ & x^2+y^2=25. \end( aligned)\right.$$

To solve this system, let's immediately indicate that $\lambda\neq -1$. Why $\lambda\neq -1$? Let's try to substitute $\lambda=-1$ into the first equation:

$$ x+(-1)\cdot x=6; \; x-x=6; \; 0=6. $$

The resulting contradiction $0=6$ says that the value $\lambda=-1$ is invalid. Output: $\lambda\neq -1$. Let's express $x$ and $y$ in terms of $\lambda$:

\begin(aligned) & x+\lambda x=6;\; x(1+\lambda)=6;\; x=\frac(6)(1+\lambda). \\ & y+\lambda y=-8;\; y(1+\lambda)=-8;\; y=\frac(-8)(1+\lambda). \end(aligned)

I believe that it becomes obvious here why we specifically stipulated the $\lambda\neq -1$ condition. This was done to fit the expression $1+\lambda$ into the denominators without interference. That is, to be sure that the denominator is $1+\lambda\neq 0$.

Let us substitute the obtained expressions for $x$ and $y$ into the third equation of the system, i.e. in $x^2+y^2=25$:

$$ \left(\frac(6)(1+\lambda) \right)^2+\left(\frac(-8)(1+\lambda) \right)^2=25;\\ \frac( 36)((1+\lambda)^2)+\frac(64)((1+\lambda)^2)=25;\\ \frac(100)((1+\lambda)^2)=25 ; \; (1+\lambda)^2=4. $$

It follows from the resulting equality that $1+\lambda=2$ or $1+\lambda=-2$. Hence, we have two values of the parameter $\lambda$, namely: $\lambda_1=1$, $\lambda_2=-3$. Accordingly, we get two pairs of values $x$ and $y$:

\begin(aligned) & x_1=\frac(6)(1+\lambda_1)=\frac(6)(2)=3; \; y_1=\frac(-8)(1+\lambda_1)=\frac(-8)(2)=-4. \\ & x_2=\frac(6)(1+\lambda_2)=\frac(6)(-2)=-3; \; y_2=\frac(-8)(1+\lambda_2)=\frac(-8)(-2)=4. \end(aligned)

So, we got two points of a possible conditional extremum, i.e. $M_1(3;-4)$ and $M_2(-3;4)$. Find the values of the function $z$ at the points $M_1$ and $M_2$:

\begin(aligned) & z_1=z(M_1)=3^2+(-4)^2-12\cdot 3+16\cdot (-4)=-75; \\ & z_2=z(M_2)=(-3)^2+4^2-12\cdot(-3)+16\cdot 4=125. \end(aligned)

We should choose the largest and smallest values from those that we obtained in the first and second steps. But in this case the choice is small :) We have:

$$z_(min)=-75; \; z_(max)=125. $$

Answer: $z_(min)=-75; \; z_(max)=125$.

In this article I will talk about algorithm for finding the largest and smallest value function, minimum and maximum points.

From theory, we will definitely need derivative table And differentiation rules. It's all in this board:

Algorithm for finding the largest and smallest values.

I find it easier to explain specific example. Consider:

Example: Find the largest value of the function y=x^5+20x^3–65x on the segment [–4;0].

Step 1. We take the derivative.

Y" = (x^5+20x^3–65x)" = 5x^4 + 20*3x^2 - 65 = 5x^4 + 60x^2 - 65

Step 2 Finding extremum points.

extremum point we name such points at which the function reaches its maximum or minimum value.

To find the extremum points, it is necessary to equate the derivative of the function to zero (y "= 0)

5x^4 + 60x^2 - 65 = 0

Now we solve this biquadratic equation and the found roots are our extremum points.

I solve such equations by replacing t = x^2, then 5t^2 + 60t - 65 = 0.

Reduce the equation by 5, we get: t^2 + 12t - 13 = 0

D = 12^2 - 4*1*(-13) = 196

T_(1) = (-12 + sqrt(196))/2 = (-12 + 14)/2 = 1

T_(2) = (-12 - sqrt(196))/2 = (-12 - 14)/2 = -13

We make the reverse substitution x^2 = t:

X_(1 and 2) = ±sqrt(1) = ±1
x_(3 and 4) = ±sqrt(-13) (we exclude, there cannot be negative numbers under the root, unless of course we are talking about complex numbers)

Total: x_(1) = 1 and x_(2) = -1 - these are our extremum points.

Step 3 Determine the largest and smallest value.

Substitution method.

In the condition, we were given the segment [b][–4;0]. The point x=1 is not included in this segment. So we don't consider it. But in addition to the point x=-1, we also need to consider the left and right borders of our segment, that is, the points -4 and 0. To do this, we substitute all these three points into the original function. Notice the original one is the one given in the condition (y=x^5+20x^3–65x), some start substituting into the derivative...

Y(-1) = (-1)^5 + 20*(-1)^3 - 65*(-1) = -1 - 20 + 65 = [b]44
y(0) = (0)^5 + 20*(0)^3 - 65*(0) = 0
y(-4) = (-4)^5 + 20*(-4)^3 - 65*(-4) = -1024 - 1280 + 260 = -2044

This means that the maximum value of the function is [b]44 and it is reached at the points [b]-1, which is called the maximum point of the function on the segment [-4; 0].

We decided and got an answer, we are great, you can relax. But stop! Don't you think that counting y(-4) is somehow too complicated? In conditions of limited time, it is better to use another method, I call it like this:

Through intervals of constancy.

These gaps are found for the derivative of the function, that is, for our biquadratic equation.

I do it in the following way. I draw a directional line. I set the points: -4, -1, 0, 1. Despite the fact that 1 is not included in the given segment, it should still be noted in order to correctly determine the intervals of constancy. Let's take some number many times greater than 1, let's say 100, mentally substitute it into our biquadratic equation 5(100)^4 + 60(100)^2 - 65. Even without counting anything, it becomes obvious that at the point 100 the function has plus sign. This means that for intervals from 1 to 100 it has a plus sign. When passing through 1 (we go from right to left), the function will change sign to minus. When passing through the point 0, the function will retain its sign, since this is only the boundary of the segment, and not the root of the equation. When passing through -1, the function will again change sign to plus.

From theory, we know that where the derivative of the function is (and we drew this for it) changes sign from plus to minus (point -1 in our case) function reaches its local maximum (y(-1)=44 as calculated earlier) on this segment (this is logically very clear, the function has ceased to increase, since it reached its maximum and began to decrease).

Accordingly, where the derivative of the function changes sign from minus to plus, achieved local minimum of a function. Yes, yes, we also found the local minimum point, which is 1, and y(1) is the minimum value of the function on the interval, let's say from -1 to +∞. Please note that this is only a LOCAL MINIMUM, that is, a minimum on a certain segment. Since the actual (global) minimum function will reach somewhere there, in -∞.

In my opinion, the first method is simpler theoretically, and the second one is simpler in terms of arithmetic operations, but much more difficult in terms of theory. After all, sometimes there are cases when the function does not change sign when passing through the root of the equation, and indeed you can get confused with these local, global maxima and minima, although you will have to master it well anyway if you plan to enter a technical university (and for what else to give profile exam and solve this problem). But practice and only practice will teach you how to solve such problems once and for all. And you can train on our website. Here .

If you have any questions, or something is not clear, be sure to ask. I will be happy to answer you, and make changes, additions to the article. Remember we are making this site together!

Let's see how to explore a function using a graph. It turns out that looking at the graph, you can find out everything that interests us, namely:

function scope
function range
function zeros
periods of increase and decrease
high and low points
the largest and smallest value of the function on the interval.

Let's clarify the terminology:

Abscissa is the horizontal coordinate of the point.
Ordinate- vertical coordinate.
abscissa- the horizontal axis, most often called the axis.
Y-axis- vertical axis, or axis.

Argument is an independent variable on which the values of the function depend. Most often indicated.
In other words, we ourselves choose , substitute in the function formula and get .

Domain functions - the set of those (and only those) values of the argument for which the function exists.
Denoted: or .

In our figure, the domain of the function is a segment. It is on this segment that the graph of the function is drawn. Only here this function exists.

Function range is the set of values that the variable takes. In our figure, this is a segment - from the lowest to the highest value.

Function zeros- points where the value of the function is equal to zero, i.e. . In our figure, these are the points and .

Function values are positive where . In our figure, these are the intervals and .
Function values are negative where . We have this interval (or interval) from to.

The most important concepts - increasing and decreasing function on some set. As a set, you can take a segment, an interval, a union of intervals, or the entire number line.

Function increases

In other words, the more , the more , that is, the graph goes to the right and up.

Function decreases on the set if for any and belonging to the set the inequality implies the inequality .

For a decreasing function, a larger value corresponds to a smaller value. The graph goes right and down.

In our figure, the function increases on the interval and decreases on the intervals and .

Let's define what is maximum and minimum points of the function.

Maximum point- this is an internal point of the domain of definition, such that the value of the function in it is greater than in all points sufficiently close to it.
In other words, the maximum point is such a point, the value of the function at which more than in neighboring ones. This is a local "hill" on the chart.

In our figure - the maximum point.

Low point- an internal point of the domain of definition, such that the value of the function in it is less than in all points sufficiently close to it.
That is, the minimum point is such that the value of the function in it is less than in neighboring ones. On the graph, this is a local “hole”.

In our figure - the minimum point.

The point is the boundary. It is not an interior point of the domain of definition and therefore does not fit the definition of a maximum point. After all, she has no neighbors on the left. In the same way, there can be no minimum point on our chart.

The maximum and minimum points are collectively called extremum points of the function. In our case, this is and .

But what if you need to find, for example, function minimum on the cut? In this case, the answer is: Because function minimum is its value at the minimum point.

Similarly, the maximum of our function is . It is reached at the point .

We can say that the extrema of the function are equal to and .

Sometimes in tasks you need to find the largest and smallest values of the function on a given segment. They do not necessarily coincide with extremes.

In our case smallest function value on the interval is equal to and coincides with the minimum of the function. But its largest value on this segment is equal to . It is reached at the left end of the segment.

In any case, the largest and smallest values of a continuous function on a segment are achieved either at the extremum points or at the ends of the segment.

A miniature and rather simple task of the kind that serves as a lifeline for a floating student. In nature, the sleepy realm of mid-July, so it's time to settle down with a laptop on the beach. Played early in the morning sunbeam theory in order to soon focus on practice, which, despite its claimed lightness, contains glass fragments in the sand. In this regard, I recommend conscientiously consider a few examples of this page. To solve practical tasks, you need to be able to find derivatives and understand the material of the article Intervals of monotonicity and extrema of a function.

First, briefly about the main thing. In a lesson about function continuity I gave the definition of continuity at a point and continuity on an interval. The exemplary behavior of a function on a segment is formulated similarly. A function is continuous on a segment if:

1) it is continuous on the interval ;
2) continuous at a point on right and at the point left.

The second paragraph deals with the so-called unilateral continuity functions at a point. There are several approaches to its definition, but I will stick to the line started earlier:

The function is continuous at a point on right, if it is defined at a given point and its right-hand limit coincides with the value of the function at a given point: . It is continuous at the point left, if defined at a given point and its left-hand limit is equal to the value at that point:

Imagine that the green dots are the nails on which the magic rubber band is attached:

Mentally take the red line in your hands. Obviously, no matter how far we stretch the graph up and down (along the axis), the function will still remain limited- a hedge above, a hedge below, and our product grazes in a paddock. Thus, a function continuous on a segment is bounded on it. In the course of mathematical analysis, this seemingly simple fact is stated and rigorously proved Weierstrass' first theorem.... Many people are annoyed that elementary statements are tediously substantiated in mathematics, but there is important meaning. Suppose a certain inhabitant of the terry Middle Ages pulled the graph into the sky beyond the limits of visibility, this was inserted. Before the invention of the telescope, the limited function in space was not at all obvious! Indeed, how do you know what awaits us beyond the horizon? After all, once the Earth was considered flat, so today even ordinary teleportation requires proof =)

According to second Weierstrass theorem, continuous on the segmentfunction reaches its exact top edge and his exact bottom edge .

The number is also called the maximum value of the function on the segment and denoted by , and the number - the minimum value of the function on the segment marked .

In our case:

Note : in theory, records are common .

Roughly speaking, the greatest value is located where the most high point graphics, and the smallest - where is the lowest point.

Important! As already pointed out in the article on extrema of the function, the largest value of the function And smallest function value – NOT THE SAME, What function maximum And function minimum. So, in this example, the number is the minimum of the function, but not the minimum value.

By the way, what happens outside the segment? Yes, even the flood, in the context of the problem under consideration, this does not interest us at all. The task involves only finding two numbers and that's it!

Moreover, the solution is purely analytical, therefore, no need to draw!

The algorithm lies on the surface and suggests itself from the above figure:

1) Find the function values in critical points, that belong to this segment.

Catch one more goodie: there is no need to check a sufficient condition for an extremum, since, as just shown, the presence of a minimum or maximum not yet guaranteed what is the minimum or maximum value. The demo function reaches its maximum and by the will of fate the same number is highest value functions on the interval . But, of course, such a coincidence does not always take place.

So, at the first step, it is faster and easier to calculate the values of the function at critical points belonging to the segment, without bothering whether they have extrema or not.

2) We calculate the values of the function at the ends of the segment.

3) Among the values of the function found in the 1st and 2nd paragraphs, we select the smallest and most big number, write down the answer.

We sit on the shore of the blue sea and hit the heels in shallow water:

Example 1

Find the largest and smallest values of a function on a segment

Solution:
1) Calculate the values of the function at critical points belonging to this segment:

We calculate the value of the function in the second critical point:

2) Calculate the values of the function at the ends of the segment:

3) "Bold" results were obtained with exponentials and logarithms, which significantly complicates their comparison. For this reason, we will arm ourselves with a calculator or Excel and calculate the approximate values, not forgetting that: